Derive relation between current and drift velocity. Why drift velocity is so slow.

Drift Velocity and Current — I = nAve

Question

Derive the relation between electric current $I$ and drift velocity $v_d$ for a conductor. Also explain why drift velocity is so small (around $10^{-4}$ m/s) even though electricity “travels” at nearly the speed of light.

Solution — Step by Step

Consider a conductor of cross-sectional area $A$ , with $n$ free electrons per unit volume (number density). Each electron carries charge $e$ .

We want to find how much charge crosses any cross-section in one second.

When an electric field is applied, electrons drift with velocity $v_d$ . In time $t$ , an electron travels a distance $v_d \cdot t$ .

So all electrons within a cylinder of length $v_d \cdot t$ and cross-section $A$ will cross our reference plane in time $t$ :

\text{Volume} = A \cdot v_d \cdot t

Number of electrons in this cylinder:

N = n \cdot A \cdot v_d \cdot t

Total charge flowing across the section in time $t$ :

Q = N \cdot e = n A v_d e \cdot t

Current is charge per unit time:

I = \frac{Q}{t} = nAv_d e

This is our result. Rearranging for drift velocity:

v_d = \frac{I}{nAe}

Take a copper wire carrying 1 A, with $A = 1 \text{ mm}^2 = 10^{-6} \text{ m}^2$ and $n \approx 8.5 \times 10^{28} \text{ m}^{-3}$ :

v_d = \frac{1}{8.5 \times 10^{28} \times 10^{-6} \times 1.6 \times 10^{-19}} \approx 7.4 \times 10^{-5} \text{ m/s}

That’s less than 0.1 mm per second. Drift velocity is extremely small.

Why This Works

The key insight is that $n$ (free electron density) in metals is enormous — around $10^{28}$ to $10^{29}$ per cubic metre. Even though each electron barely moves, the sheer number of them crossing the section every second adds up to a significant charge flow.

Think of it like a pipe packed with balls. You push one end, and the ball at the other end moves almost instantly — not because each ball travelled fast, but because the disturbance (pressure wave) propagated at high speed. In a conductor, the electric field propagates at close to the speed of light ( $\sim 3 \times 10^8$ m/s), which is why your bulb lights up instantly. But the electrons themselves shuffle along at a snail’s pace.

This is a favourite conceptual question in board exams and JEE. The “speed of electricity” and drift velocity are two entirely different things — the field propagates fast, the electrons drift slow.

Alternative Method

For MCQs where $n$ isn’t given directly, remember the ratio form. If two wires of the same material have areas $A_1$ and $A_2$ carrying the same current $I$ , then:

\frac{v_{d1}}{v_{d2}} = \frac{A_2}{A_1}

Thinner wire → larger drift velocity. This appeared as a one-liner in JEE Main 2024 Shift 1.

You can also arrive at the formula by thinking in terms of current density $J$ :

J = \frac{I}{A} = nev_d

This form is more general — it works even when the cross-section isn’t uniform. For a conductor with varying area, $J$ changes but $I$ stays constant, so $v_d$ adjusts with $1/A$ .

Common Mistake

Students often write $v_d = \frac{I}{ne}$ forgetting the area $A$ . The formula is $I = nAv_de$ , so $v_d = \frac{I}{nAe}$ . Without $A$ , the units don’t even work out: the left side is m/s, but $\frac{I}{ne}$ gives m³/s. If your units are off, the formula is wrong — always check dimensions before moving on.

A second slip: confusing $n$ (number density, unit m $^{-3}$ ) with $N$ (total number of electrons). Use $n$ in the formula — it’s an intensive property of the material, not of the specific piece of wire you’re given.

\boxed{I = nAv_de}

Where:

$n$ = free electron number density (m⁻³)
$A$ = cross-sectional area (m²)
$v_d$ = drift velocity (m/s)
$e$ = charge of electron ( $1.6 \times 10^{-19}$ C)

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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